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Math Help - dual map dimension

  1. #1
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    Joined
    Dec 2009
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    11

    dual map dimension

    F: field
    V,W: F vector spaces
    V*,W*: dual spaces (Hom(V,F);Hom(W,F))
    φ : V -> W linear map
    φ *:W*->V*
    (λ: W -> F) |-> (λ∘ φ: V -> F)

    Proof that dim(im(φ))=dim(im(φ*))
    I have actually no idea how to proof it. I tried to use the rank-nullity theorem.
    We know:
    dim(V*)=dim(V)*dim(F)=dim(V)
    dim(W*)=dim(W)*dim(F)=dim(W)

    dim(V)=dim(ker(φ))+dim(im(φ))
    dim(W)=dim(ker(φ*))+dim(im(φ*))
    (r-n t)

    Additionally:
    dim(im(φ))<=dim(W)
    dim(im(φ*))<=dim(V)

    But how can I use these to show that the images have the same dimension? Maybe it is not possible how I tried it.

    Is it somehow possible to show that there is an isomorphism between im(φ) and im(φ*)?

    Has it something to do with the base of a dual space?
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  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Hi,
    you can't just use these formulas, you need more.

    For instance, you can start by looking at \ker \phi^*. It consists of all \lambda\in W^* such that \lambda\circ\phi=0. This condition is equivalent to {\rm Im} \phi\subset \ker \lambda: \lambda has to be zero on {\rm Im} \phi, and may be equal to anything on a supplementary space. Thus the dimension of the set of such \lambda\in W^* is \dim W - \dim {\rm Im}\phi. But we also have \dim \ker \phi^* = \dim W - \dim {\rm Im }\phi^*, hence we conclude \dim {\rm Im} \phi=\dim {\rm Im} \phi^*.

    The statement about " \lambda has to be zero on {\rm Im} \phi, and may be equal to anything on a supplementary space" and the computation of the dimension are made rigorous using bases or dual bases. For instance you can take a base of W that starts with a base of {\rm Im} \phi, so that \lambda\in\ker \phi^* iff it equals 0 on the first vectors of the basis. From there you can deduce that \ker\phi^* is spanned by the elements of the dual base that correspond to vectors not in \ker\phi^*. (I know this is not very clear, that's just the rough idea, I let you try to find by yourself)
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